Explain how you know 21/100 is greater than 1/5

1 answer:

5 0

This problem can be solved using equivalent fractions. The first step in resolving this problem is to realize that fractions are best compared when they both have the same denominator. In this case, I will choose to make that common denominator 100. There is no need to rewrite the fraction $\frac{21}{100}$ as the denominator for this fraction is already 100. The fraction $\frac{1}{5} =\frac{20}{100}$ . This is achieved by multiplying both the numerator and denominator by 20. Now that the two fractions have the same denominator, we can easily see that $\frac{21}{100}$ is greater than $\frac{1}{5}$ because it is greater than its equivalent fraction.

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What is the y-intercept of the line shown below? A. 5 B. -5 C. 2 D. -2

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Answer: -2

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F(t)=t^2+19t+60 What are the zeros of the function and what is the vertex of the parabola?

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Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$